## Beautiful Books

This is an (ever-growing and ever-changing) list of books, useful for school and college mathematics students. If you are working toward Math Olympiad, I.S.I., C.M.I. entrance programs or intense college mathematics, these books may prove to be your best friend.

If you are taking a Cheenta Advanced Math Program, chances are that you will referred to use this post.

## I.S.I. Entrance Interview Problems

1. a and b are two numbers having the same no. of digits and same sum of digits (=28). Can one be a multiple of the other? a is not equal to b. (courtesy Abhra Abir Kundu)
2. Is $$e^x-sinx$$ a polynomial ? (courtesy Tias Kundu)
3. Find the number of onto function from set A containing n elements to set B containing m elements (m<n) (courtesy Tias Kundu)
4. If a+b+c=30,  how many (a,b,c) tuples possible (a,b,c all non-negative). (courtesy Saikat Palit)
5. Can sin(x) be expressed as a polynomial in x? (courtesy Soumik Bhattacharyya)
6. Integers 1-64 are placed in a 8X8 chessboard. How many ways are there to place them such that all numbers in the 1st row and column are in AP? (courtesy Soumik Bhattacharyya)

## Math and Basic Science in India after school

We observe a renewed interest in basic science and mathematics in India after decades of engineering mania. This is a welcome change. The top tier students, in increasing numbers, are opting for such courses.

In this article we present some of the institutes which offer world class education in basic science after school.

## Initiating a child into the world of Mathematical Science

“How do I involve my son in challenging mathematics? He gets good marks in school tests but I think he is smarter than school curriculum.”

“My daughter is in 4th grade. What competitions in mathematics and science can she participate? How do I help her to perform well in those competitions?”

“I have a 6 years old kid. He hates math. How do I change that?”

We often get queries and requests like these from parents around the world. Literally. In fact the first one came from Oregon, United States, second one from Cochin, India and last one from Singapore.

We have created this article to answer these kind of questions and help you to help your children.

# What is RMO

RMO or Regional Math Olympiad is the first round of mathematics contest (in India) leading to the prestigious International Mathematics Olympiad. It is held in the month of December (first Sunday of December).

The test is conducted in each of the 19 regions of India. From each region about 30 students are selected for the next level (Indian National Math Olympiad). This next level (INMO) is held in February. About 600 students appear in this INMO. Among these 600 students 35 are selected for a one month long camp (International Math Olympiad Training Camp) held in summer in Mumbai.

## Angle between Force and Momentum

A particle moves in the X-Y plane under the influence of a force such that its linear momentum p(t)=A[icos(kt)-jsin(kt)] where A and k are constants. Find the angle between the force and the momentum.

Solution:

In the given problem, linear momentum p(t)=A[icos(kt)-jsin(kt)]

We know, force$$F=\frac{dp}{dt}$$
Differentiating p(t) with respect to t, we get

$$F=\frac{dp}{dt}=A[-iksin(kt)-jkcos(kt)]$$
Taking dot product of F and p,
$$F.p=-ksin(kt)cos(kt)+kcos(kt)sin(kt) \Rightarrow 0$$
Since, the dot product is zero, we can conclude that $$cos\theta=0.$$ where θ is the angle between the force and momentum.
Hence, it is implied that $$\theta=90^\circ$$
Thus, the angle between the force and momentum will be ninety degrees.

## Motion in an Electric Field

A particle moves rectilinearly in an electric field E=E0-ax where a is a positive constant and x is the distance from the point where the particle is initially at rest. Let the particle have a specific charge q/m.

Find:

(I) the distance covered by the particle till the moment at which it once again comes to rest, and

(II) acceleration of the particle at this moment.

Solution:

A particle moves rectilinearly in an electric field $$E=E_0-ax$$ where a is a positive constant and x is the distance from the point where the particle is intially at rest.
The particle has a specific charge q/m.
Now,

$$F=q(E_0-ax)$$
$$or, a = \frac{q(E_o-ax)}{m}$$
At x=0,

$$a=\frac{qE_0}{m}$$
Particle will move in the x direction
$$\frac{vdv}{dx}=a$$

$$v\frac{dv}{dx}=\frac{q(E_0-ax))}{m}$$
$$vdv=\frac{q(E_0-ax)}{m}dx$$
$$\int_{0}^{0} vdv=\int_{0}^{x_0} \frac{q(E_0-ax)}{m}dx$$
$$0=\frac{q(E_0x-\frac{ax^2}{2})}{m}$$
Now, $$v=0, x=x_0$$
Hence,

$$E_0x_0=a\frac{x_0^2}{2}$$
$$x_0=\frac{2E_0}{a}$$
Distance covered by the particle before coming to rest =
$$\frac{2E_0}{a}$$
Acceleration before coming to rest will be
$$a=\frac{-qE_0}{m}$$
The direction of the particle will be towards the negative x-axis.

## Roots of an equation (TIFR 2013 problem 3)

Problem

The equation $$x^3+3x-4=0$$ has exactly one real root.

Discussion

Hint: Try differentiating. Is the function monotonic?

Note that cubic equation has at least one real root (fundamental theorem of algebra provides for three roots counting repetitions, real or complex; for equations with real coefficients, the complex roots appear in conjugate pairs. Hence a cubic equation with real coefficients must have at least one real root).

Now differentiating $$P(x) = x^3 + 3x – 4$$ we have the following: $$\frac {d} {dx} P(x) = 3x^2 + 3$$

What can you say about this derivative?

Clearly it is always positive (perfect square plus three is at least three or more).

Hence P(x) is monotonically increasing. It will ‘cut’ the x-axis only at one point.

Therefore P(x) has exactly one real root.

Remark:

In this problem, first we established P(x) has at least one real root. Then we proved it has exactly one.  Both steps are crucial.

## Power and Acceleration

Assume that a constant power P is supplied to an electric train and it is fully used in accelerating the train. Obtain relation giving the velocity of the train and distance travelled by it as functions of time.

Solution:

We know, power (P)= force(F)*velocity (v).

Therefore,

m dv/dt=P/v

or, vdv=P/m dt

Integrating both sides, we have

∫vdv= P/m ∫dt

v2/2=P/m t+c

where c is a constant of integration.

Now, at t=0, v=0 so c=0.

Therefore,

Or, v = √(2Pt/m)……… (i)

We know, v= dx/dt

So, from equation (i)

dx/dt= √(2Pt/m)

Integrating both sides,

∫dx= √(2Pt/m) dt

or, x= (2/3)√(2P/m)t3/2

## Instantaneous Acceleration

Velocity-displacement curve of a particle moving in a straight line is as shown

• 2m/s2
• 5 m/s2
• 1 m/s2
• Zero

Discussion:

Acceleration a=v dv/ds

From, the graph, we can see a=vtanθ

tanθ=1/4

Putting the values, we get acceleration a= 1m/s2

## Projectile Inside a Liquid

A body of mass m is projected inside a liquid at an angle θ0 with horizontal at an initial velocity v0. If the liquid develops a velocity dependent force F= -kv where k is a positive constant, determine the x and y components of the velocity at any instant.

Solution:

A body of mass m is projected inside a liquid at an angle θ0 with horizontal at an initial velocity v0. The liquid develops a velocity dependent force F= -kv where k is a positive constant.

Hence,

m dv/dt= -kv

or, dv/dt= -k/m v

or, dv/v= -k/m dt

Integrating both sides,

∫dv/v = -k/m ∫dt

ln|v|= -kt/m+c (where c is a constant of integration)….. (i)

Now, for the x component of velocity,

ln vx=  -kt/m+ c

From the given problem, we have

vx=v0 cosθ0 at t=0

Applying the above condition in eqn.(i), we get

ln (vx/ v0 cosθ0 )= -kt/m

or vx=v0cosθ0e-kt/m

For the y component, we have to consider the acceleration due to gravity g.

Hence,

m dvy/dt= -kvy-mg

or, dvy/(kvy+mg)= -k/m dt

Integrating both sides,

ln|kvy+mg|=-kt/m+c

At t=0, vy=v0sinθ0

Hence,

kvy+mg=(kv0sinθ0+mg) e-kt

## Motion under Constant Gravity

A person throws vertically up n balls per second with the same velocity. He throws a ball whenever the previous one is at its highest point. The height to which the balls rise is

(a) g/n2

(b) 2gn

(c) g/2n2

(d) 2gn2

Solution:

We know v=u-gt. v is zero at the highest point. Time t taken by one ball to reach maximum height is 1/n.

Hence, we have

u=gt

or, u=g/n……. (i)

now, from the relation v2=u2-2gh. Again, v=0

u2=2gh……. (ii)

Putting the value of u from equation (i) in above relation (ii), we get

h=g/2n2.

## Velocity and Acceleration

A particle is moving in positive x-direction with its velocity varying as v= α√x. Assume that at t=0, the particle was located at x=0. Determine the

• the time dependence of velocity
• acceleration
• the mean velocity of the particle averaged over the time that the particle takes to cover the first s metres of the path.

Discussion:

v=α√x.

Squaring both sides

v22x

=2(α2/2)x

Acceleration= α2/2

The initial velocity u is therefore zero and the acceleration is constant.

• v=u+at= (α2/2)t
• The acceleration= α2/2
• V=α√s

Average velocity=(0+v)/2=α√s/2

## Understanding the Infinitesimal

Understanding the Infinitesimal
Cheenta Notes in Mathematics

Adding infinitely many positive quantities, you may end up having something finite. Greeks did not understand this very well. Archimedes had some idea. Kerala school of mathematics under the leadership of Madhavacharya made real progress in refining this notion. They created the necessary groundwork for the advent of calculus later in Germany and Britain.

Indian Statistical Institute’s 2017 entrance for B.Stat and B.Math had a very simple problem based on infinitesimals. Adding little things, can you end up having more than ‘little’?

Here is the problem:

Given f : ℝ → ℝ be a continuous function such that for any two real numbers x and y,
|f(x) – f(y)| ≤ 7|x-y|201

Then:

(A) f(101) = f(202) + 8;
(B) f(101) = f(201) +1;
(C) f(101) = f(200) + 2;
(D) None of the above;

Before we solve the actual problem, lets have a fun detour. What if f is differentiable (the problem does not say that) ? Then take y = x + δ where δ > 0 Clearly, by the given condition:

| f(x) – f(x + δ) | ≤ 7 |x – (x + δ) |201 = 7δ201
⇒ |f(x) – f(x+δ) |/δ ≤ 7 δ200
⇒ limδ → 0 |f(x) – f(x+δ) |/δ ≤ limδ → 0  7δ200 =  0
⇒ |f'(x)| = 0

That means, if f is differentiable, then it’s derivative is 0, or in other words it is constant function. In that case f(x) = f(y) = c (a constant for all x and y). Hence the none of the first three options would hold. (Interestingly enough this is sufficient to choose (D) as the correct option as differentiable functions are an important subclass of continuous functions).

However we cannot assume differentiability as it is not mentioned in the problem. But now we have a hunch! We are already guessing that may be f is not changing much.

Suppose we want to check:
| f(101) – f(202) |

Chop off the distance between 101 to 202 into intervals of 0.1 unit long. There are 1010 such intervals (this is the ‘adding the little’ part). Why did I choose 0.1 length? Well, that is because it is a fractional length and raising a fraction to large powers will make it even smaller.

Now note that:

|f(101) – f(202)|
= |f(101) – f(101.1) + f(101.1) – f(101.2) + … + f(201.9) – f(202)|
≤  |f(101) – f(101.1)| + |f(101.1) – f(101.2)| + … + |f(201.9) – f(202)|
≤ 7|101-101.1|201 + … + 7|201.9-202|201
= 7 ( 0.1201 + … + 0.1201
= 7 × 1010 × 0.1201
= 7070/10201

But that implies |f(101) – f(202)| is much smaller 1 let alone 8. Using this same technique you can make |f(101)  –  f(202| smaller than 1 and f(101) – f(200)| smaller than 2. Hence the answer is option (D).

More interestingly, can you make the difference between f(x) and f(y) arbitrarily small? If you can do that then f(x) will be a constant function! This is something for you to think about this week.

Here is a possible way to think about it:

Taking smaller intervals. For example, for integers x and y chop off |x-y| into intervals of length 1/n. There will be |x-y|/(1/n) = n|x-y| intervals. Then you can use the above algorithm to compute:

|f(x) – f(y)|
≤ 7 × n|x-y|/n201
= 7 × |x-y|/n200

No matter how large |x-y| is, (by archimedean property) we can find a positive integer k such that

|x-y| < k × n

Try to finish it off from here. Let me know if you get something.

On another note:

• RMO registration begins on May 20th. Keep an eye for that.
• Cheenta I.S.I. & C.M.I. Entrance, Math and Physics Olympiad batches have commenced. If you are Cheenta student, you may need class access for resources, notes and tests. Let us know if you do not have access to the portal already.
• Raychaudhury Mathematics Award registration is open. Respond to this email or call us to register.

All the best!

Ashani Dasgupta
Cheenta

vidya dadati vinayam

## Dimensional Analysis

The distance travelled by an object is given by x=(at+bt2)/(c+a) where t is time and a,b,c are constants. The dimension of b and c respectively are:

• [L2T-3], [LT-1]
• [LT-2], [LT-1]
• [LT-1], [L2T1]
• [LT-1], [LT2]

Solution:

The dimension of x is L. Hence, the dimension of a must be LT-1.

Dimension of c and a are the same.

Dimension of b therefore is L2T-3.